Let $$\phi: M_n (\mathbb{C})\otimes M_m (\mathbb{C})\to M_{nm} (\mathbb{C})$$ $$ \phi({A}\otimes{B}) = \begin{bmatrix} a_{11} {B} & \cdots & a_{1n}{B} \\ \vdots & \ddots & \vdots \\ a_{m1} {B} & \cdots & a_{mn} {B} \end{bmatrix}, $$ It is obvious $\phi$ is one to one & onto.
Question: For $C\in M_{nm} (\mathbb{C})$, what is $\phi^{-1}(C)$
Well, nothing special: write $C \in M_{mn}(\mathbb C)$. Let $C_{ab} \in M_{mn} (\mathbb C)$, where $a,b \in \{1, 2, \cdots, n\}$ so that
$$(C_{ab})_{ij} = \begin{cases} C_{ij} & \text{if } (a-1)m +1 \le i\le am, (b-1)m+1\le j\le bm,\\ 0 & \text{otherwise.}\end{cases}$$
Abusing notations, we also consider $C_{ab} \in M_m(\mathbb C)$. Then
$$C = \sum_{a,b =1} ^{n} C_{ab} = \sum_{a,b =1} ^{n} \phi(E_{ab} \otimes C_{ab}) = \phi \left(\sum_{a,b =1} ^{n} E_{ab} \otimes C_{ab}\right),$$
where $E_{ab} \in M_{n}(\mathbb C)$ is the matrix with $1$ in the $(a, b)$ entries and $0$ otherwise. Thus
$$\phi^{-1}(C) = \sum_{a,b = 1}^{n} E_{ab} \otimes C_{ab}.$$